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ale4655 [162]
3 years ago
7

Suppose 8 people want to communicate with each other using public key encryption. The communication between any pair of them is

visible to all of them but no other person other than the pair should be able to decode their communication. How many public and private keys are required for this group?
Computers and Technology
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

The answer is "64"

Explanation:

There has been 1 Private Key 1 Public Key for every person. The private key is used for other app messages. All else uses the Public Key to encrypt messages with that user. Those keys are numerically related. Even if you have 5 users, you will be given 5 Personal keys and 5 Public keys.

  • All user will have a duplicate of other public keys, which indicates (n*n-1) copies of n public keys on different systems to make sure shared contact between all users, plus n private keys.
  • The unique key count is 2n, with n^2 distributed private and public keys on different systems.
  • 8 people,  where 8 is a number, that is equal to n so, n^2 = 8^2 =8 \times 8 = 64
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Write a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters.
vichka [17]

Answer:

Explanation:

The python code for the question is attached in the image below.

The objective here is to write a code in python with a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters. Otherwise, it should return False.

SEE BELOW FOR THE CODE.

alphabet = "abcdefghijklmnopqrstuvwxyz"

test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]

test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]

def histogram(s):

   d = dict()

   for c in s:

       if c not in d:

           d[c]=1

       else:

           d[c] += 1

   return d

def has_duplicates():

   for string in test_dups:

       dictionary=histogram(string)

       duplicates=False

       for ch in dictionary.keys():

           if dictionary[ch] > 1:

               duplicates=True

               break

       if duplicates==True:

           print(string,"has duplicates")

       else:

           print(string,"has no duplicates")

def missing_letters(string):

   answer=""

   for ch in alphabet:

       if ch not in string:

           answer=answer+ch

   return answer

print("__________________________________________")

print(" Calling has_duplicates() function")

print("__________________________________________")

has_duplicates()

print("\n______________________________________________")

print("Calling missing_letters() function in for loop")

print("______________________________________________")

for i in test_miss:

   answer=missing_letters(i)

   if len(answer)>=1:

       print(i,"is missing letters",answer)

   else:

       print(i,"uses all the letters")

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Answer:

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Explanation:

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